# Deriving E=mc^2 via Units?

*Most recent answer: 05/10/2015*

- Merton Hale (age 69)

Belgium

Those dimensional analysis arguments are wonderful but not magic. If there are two possible velocties that might appear in the energy formula, dimensional arguments can never tell you what combination of the two appears. So v^{2}, c^{2}, cv, etc. are all equivalent dimensionally.

Wit regard to Maxwell's equations, they can and did give a relation E=pc between the energy and momentum of an electromagnetic wave traveling in one direction. If you write the momentum p as a product of the velocity and an inertial mass m, then that gives E=mc^{2}. That equation in effect defines m, for light. Maxwell's equations themselves don't say that it should work for anything else. They also don't say whether that m has any significance other than just providing a way of re-writing E=pc. Other arguments are needed to show that this m is what shows up as a source of gravity.

Mike W.

*(published on 05/10/2015)*